Characteristic and Continuous X-rays, Moseley's Law (JEE Advanced)

“Characteristic and continuous X-rays, Moseley's law”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, Gamma rays).

JEE Main covers X-rays only as part of the electromagnetic spectrum without going into production mechanisms or Moseley's law. For Advanced, you need to understand the physics of X-ray generation, calculate cutoff wavelength, and apply Moseley's law to find atomic numbers.

• Continuous X-rays arise from deceleration of electrons hitting the target; maximum energy equals eV (accelerating voltage times electron charge), giving a minimum wavelength cutoff
• Characteristic X-rays are emitted when electrons knock out inner-shell electrons; the resulting transitions (like L to K shell) produce specific spectral lines called Kα, Kβ, etc.
• Moseley found that √ν is proportional to (Z − b), where Z is atomic number and b is a screening constant; this helped establish atomic number as the fundamental property of elements
• The cutoff wavelength λ_min depends only on accelerating voltage, not on target material; characteristic wavelengths depend on the target element
• Kα X-rays correspond to L → K transition, Kβ to M → K transition; Kβ has higher energy (shorter wavelength) than Kα
• Intensity of continuous spectrum increases with atomic number of target and with tube current

• λ_min = hc / eV (cutoff wavelength)
• λ_min (in Å) = 12400 / V (when V is in volts)
• Moseley's Law: √ν = a(Z − b), where a and b are constants
• For Kα line: b ≈ 1, so √ν = a(Z − 1)
• Energy of Kα photon: E_Kα = 13.6 × (Z − 1)² × (1/1² − 1/2²) eV = 10.2(Z − 1)² eV
• Frequency ν = c/λ, Energy E = hν = hc/λ
• 1 eV = 1.6 × 10⁻¹⁹ J, hc = 1240 eV·nm = 12400 eV·Å

• Confusing cutoff wavelength (property of voltage) with characteristic wavelength (property of target element)
• Forgetting the screening constant b in Moseley's law; for K-series, b = 1, not 0
• Using wrong transition for Kα; it is L → K (n = 2 to n = 1), not M → K
• Mixing up energy and wavelength relationships; higher energy means shorter wavelength, so Kβ has shorter wavelength than Kα

Question. An X-ray tube operates at 40 kV. Find the cutoff wavelength. If the target is molybdenum (Z = 42), estimate the wavelength of Kα X-rays. (Use hc = 12400 eV·Å)

Solved twice independently; both solutions agree.